The tangents to the curve y = (x -2)^2 -1 at its points of inters… - Vidyadip

MCQ

The tangents to the curve $y = (x -2)^2 -1$ at its points of intersection with the line $x -y = 3$, intersect at the point

A
$\left( {\frac{5}{3},1} \right)$
B
$\left( {-\frac{5}{2},-1} \right)$
C
$\left( {-\frac{5}{2},1} \right)$
✓
$\left( {\frac{5}{2},-1} \right)$

✓

Answer

Correct option: D.

$\left( {\frac{5}{2},-1} \right)$

d
$x - y - 3 = 0\,\,\,\,\,\,\,\,\,\,......\left( i \right)$

will be chord of contact of parabola

Let the required point is $P\left( {{x_1},{y_1}} \right)$ chord of contact for point $P$ is

$\frac{{y + {y_1}}}{2} = x{x_1} - 4\frac{{\left( {x + {x_1}} \right)}}{2} + 3$

$y + {y_1} = 2{x_1}x - 4x - 4{x_1} + 6$

As equation $(i)$ and $(ii)$ are same line

$\frac{{2{x_1} - 4}}{1} = \frac{{ - 1}}{{ - 1}} = \frac{{ - 4{x_1} - {y_1} + 6}}{{ - 3}}$

$ \Rightarrow 2{x_1} - 4 = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4{x_1} - {y_{ 1}} + 6 = - 3$

${x_1} = \frac{5}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 10 - {y_{\kern 1pt} } + 9 = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y_1} = - 1$

Hence correct answer is $\left( {\frac{5}{2}\,, - 1} \right)$ which is option $(D)$.

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